Prove that $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$

Question

Problem. (Nguyễn Quốc Hưng) Let $0\le a,b,c\le 3;ab+bc+ca=3.$ Prove that $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$$ I have one solution but ugly, so I 'd like to find another. I will post my solution in the answer.

Answer 1

$\textbf{Solution.}$ (Khang Nguyen) We have $$(3-a)(3-b)(3-c)\ge 0 \rightarrow abc+9(a+b+c)\le 36$$ Assume that $$(b-1)(c-1)\ge 0 \rightarrow 36\ge 8a+(9+a)(b+c)\Rightarrow b+c\le \dfrac{(36-8a)}{9+a}$$ Therefore \begin{align*} \text{LHS}&=\sqrt{2a+b+c+2\sqrt{(a+b)(a+c)}}+\sqrt{b+c}\\& \le \sqrt{2a+\dfrac{36-8a}{9+a}+2\sqrt{a^2+3}}+\sqrt{\dfrac{36-8a}{9+a}}\\&\le 3+\sqrt{3}. \end{align*} Let $\dfrac{36-8a}{9+a}=x^2\, \left(1\le x\le 2\right),$ after simplify it becomes: $$ \left( {x}^{2}+8 \right) \left( \sqrt {3}+3-x \right) \ge \sqrt {{x}^{2 }+8}\sqrt {{x}^{4}+4\sqrt {3}\sqrt {7{x}^{4}-50{x}^{2}+124}-10 {x}^{2}+72}$$ Or $$\sqrt{x^2+8} \left( \sqrt {3}+3-x \right) \ge \sqrt {{x}^{4}+4\sqrt {3}\sqrt {7{x}^{4}-50{x}^{2}+124}-10 {x}^{2}+72}$$ Since $x\le 2\rightarrow \sqrt{3}+3-x\ge \sqrt{3}+3-2>0\rightarrow \text{VT}>0.$ By squaring both sides, we need to prove $$-2 \left( \sqrt {3}+3 \right) \left( \sqrt {3}{x}^{2}+{x}^{3}-6{x }^{2}-10\,\sqrt {3}+8\,x+6 \right) \ge 4\sqrt {3}\sqrt {7\,{x}^{4}-50\, {x}^{2}+124}$$ It's easy to prove $\text{LHS}>0$ and from here the inequality is equivalent to $$24\left( 2+\sqrt {3} \right) \left( {x}^{2}+8 \right) \left( 2 \sqrt {3}x+{x}^{2}+8\sqrt {3}-9x-10 \right) \left( x-2 \right) \left( x-1 \right) \ge 0,$$ One more time, we can check that this inequality is true for all $1 \le x \le 2$ and done!

Equality holds when $(a,b,c)=(3,1,0)$ and permutations.

See also here.

Answer 2

Some thoughts:

WLOG, assume $a\ge b \ge c$.

By Cauchy-Bunyakovsky-Schwarz inequality, we have $$\sqrt{a+b} + \sqrt{b + c} + \sqrt{c + a} \le \sqrt{ \left( \frac{a + b}{2} + (b+c) + \frac{c+a}{\sqrt{3}}\right)(2 + 1 + \sqrt{3})}.$$ It suffices to prove that $$\frac{a + b}{2} + (b+c) + \frac{c+a}{\sqrt{3}} \le 3 + \sqrt{3}.$$ This is true. Is there nice solutions?